Series and Progression Quiz 1

Solution:
a = 1
4th term = ar^{3} = (-8)
So r = -2
8th term = ar^{7} = 1 x (-2)^{7} = -128

Solution:
27, 81,243,…..2187

a = 27
r = \frac{a2}{a1} = \frac{81}{27} = 3
Let n = no. of terms
nth term = 2187 = ar^{(n-1)}
r^{n-1} = 81
3^{n-1} = 81
n-1 = 4
n = 5

r = 2
l= 448
Sum = 889
Let a = first term
l = 448 =ar^{n-1} =\frac{a\times 2^{n}}{2}
a \times 2^{n} = 448 \times 2 = 896
Sum =\frac{ a(r^{n} - 1)}{(r-1)} = \frac{ a(2^{n} - 1)}{(2-1)}
889 = a\times 2^{n} - a = 896 - a
a = 896 - 889 = 7

This is a GP with a=2 and r= \frac{1.6}{2} = 0.8
-1 < r <1 , So Sum = [latex]\frac{a}{1-r}[/latex]
= \frac{2}{1-0.8} = 10

Solution:
This is a GP with first term, a = 20
r =0.75 and n = 6
Length at nth hours = a x r^{n-1}
Length at 6th hours = 20 x 0.75^{6 - 1} = 4.74 cm

Solution:
Let a = first term of series.
r = \frac{1}{3}
Sum = \frac{a}{1-r} = 12
a = 12 x (1 - r) = 12 x (1 - \frac{1}{3} ) = 8
Third term = a\times r^{3 - 1} = 8 \times (\frac{1}{3})^{2} = \frac{8}{9}

Solution:
Let a, ar, ar^{2}, a^{3} be the first four terms of GP
a + ar = 16
ar^{2} + ar^{3} = 144

Dividing both, we get
\frac{ar^{2} + ar^{3}}{a + ar} = \frac{144}{16} = 9
\frac{ar^{2}(1+r)}{a(1+r)} = 9
r^{2} = 9
So r = +3 or -3

Solution:
GM of AB = \sqrt{ GM of A \times GM of B}
= \sqrt{12 \times20}
= \sqrt{240}
=15.49

Solution:
Reciprocal of every term of HP makes up an AP
Sum of 7 terms of AP = 84
\frac{7}{2} x ( 2a + 6d) = 84
Solving, a + 3d = 12
Fourth term of AP= a + 3d = 12
Fourth term of HP = \frac{1}{12}

Solution:
AP is obtained by taking reciprocal of terms of HP
Fourth term of AP = \frac{99}{7}
Seventh term of AP = \frac{85}{7}
Difference = \frac{99-85}{7} = 2
(a + 3d) - (a + 6d) = 2
d = \frac{-2}{3}
4th term of AP = a + 3d = \frac{99}{7}
a - 2 =\frac{99}{7}
a = \frac{99}{7} + 2 = \frac{113}{7}
So first term of HP = \frac{7}{113}